Calculate the enthalpy change when one mole of `HCl(g)` is dissolved in a very large amount of water at `25^(@)C`. The change in state is: `HCl(g)+arrarrH^(+)(aq)+Cl^(-)(aq)`
Given: `Delta_(f)H(HCl, g) -92 KJ mol^(-1)` and `Delta_(f)H^(@)(Cl^(-),aq)= -167 KJ mol^(-1)`

Explanation

To calculate the enthalpy change when one mole of HCl(g) is dissolved in a large amount of water at 25°C, we can use the enthalpy of formation values provided for HCl(g) and Cl⁻(aq). The reaction can be represented as:

\[ \text{HCl(g)} \rightarrow \text{H}^+(aq) + \text{Cl}^-(aq) \]

### Step-by-Step Solution:

1. **Identify the Reaction**:
The dissolution of HCl gas in water can be represented as:
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